1. Hess’s Law

2. Some enthalpy changes can not be measured directly in the lab. Hess’s Law is used to calculate these enthalpy changes Use the information provided to calculate ∆H f CH 4

3. C (gr) + 2H 2(g) CH 4(g) CO 2(g) + 2H 2 O (g) + 2O 2(g)  H c C -394 2 x  H c H 2 2 x -286 = -572  H c CH 4 -890 ∆H f CH 4

4. ∆H f CH 4 = -76kJmol -1

5. C (gr) + 2H 2(g) CH 4(g) CO 2(g) + 2H 2 O (g)  H c C -394 2 x  H c H 2 2 x -286 = -572  H c CH 4 -890 ∆H f CH 4 ∆H f CH 4 =  H c C + 2 x  H c H 2 -  H c CH 4 ∆H f CH 4 = -394 + (2 x -286) - (-890) = -76kJmol -1

6. Hess’s Law Hess’s Law states In a reaction the total enthalpy change is independent of the route

7. Hess’s Law Diagrams Write unknown equation across the top Look at info provided & complete triangle Balance equations (ignore O 2 ) Think about definitions of ∆H f & ∆H c Make sure arrows point in the correct directions Do you need to multiply any values? Write figures on arrows if this helps

8. Hess’s Law Diagrams Draw overall direction arrows Write out in full ∆H = expression without n o. In same direction as arrow + Against direction of arrow - Put numbers into ∆H = expression, keep ( ) Be very careful with signs Calculate value – check over !!!

9. Hess’s Law Diagrams Draw a Hess’s Law diagram to calculate the  H r of the coffee can reaction CaO (s) + H 2 O (l) → Ca(OH) 2 (s)  H f CaO (S) = -635  H f H 2 O (l) = -286  H f Ca(OH) 2(S) = -986

10. CaO (s) + H 2 O (l) Ca(OH) 2(s) Ca (s) + O 2(g) + H 2(g)  H f CaO -635  H f H 2 O -286  H f Ca(OH) 2 -986 ∆H r ∆H r = -  H f CaO (s) -  H f H 2 O (l) +  H f Ca(OH) 2 (s) ∆H f CH 4 = -(-635) – (-286) + (-986) = -65kJmol -1

11. Hess’s Law Hess’s Law states In a reaction the total enthalpy change is independent of the route Draw a Hess’s Law diagram to calculate the  H r of the decomposition of NaHCO 3(s) 2NaHCO 3(s) → Na 2 CO 3(s) + CO 2(g) + H 2 O (l)

12. 2NaHCO 3(s) Na 2 CO 3(s) + H 2 O (l) + CO 2(g) 2Na (s) + 3O 2(g) + H 2(g) + 2C (gr) 2 x  H f NaHCO 3 2 x -948 ∆H r ∆H r = - 2 x  H f NaHCO 3(s) +  H f Na 2 CO 3 +  H f H 2 O (l) +  H f CO 2 (g) ∆H r = -(2 x -948) + (-1131) + (-286) + (-394) = +85kJmol -1  H f CO 2 -394  H f Na 2 CO 3 -1131  H f H 2 O -286

13. Hess’s Law Draw a Hess’s Law diagram to calculate the following :  H r of the formation of propane C 3 H 8(g) 3C (gr) + 4H 2(g) → C 3 H 8(g)  H r of photosynthesis 6CO 2(g) + 6H 2 O → C 6 H 12 O 6(aq) + 6O 2(l)

14. 3C (gr) + 4H 2(g) C 3 H 8(g) 3CO 2(g) + 4H 2 O (g) + 3O 2(g) 3 x  H c C 3 x -394 = -1182 4 x  H c H 2 4 x -286 = -1144  H c C 3 H 8 -2220 ∆H f C 3 H 8 ∆H f C 3 H 8 = 3x  H c C + 4 x  H c H 2 -  H c C 3 H 8 ∆H f C 3 H 8 = (3 x -394) + (4 x -286) - (-2220) = -106kJmol -1

15. 6CO 2(g) + 6H 2 O (l) C 6 H 12 O 6 (s) + 6O 2 (g) 6C (gr) + 9O 2(g) + 6H 2(g) 6 x  H f CO 2 6 x -394 ∆H r ∆H r = - 6 x  H f CO 2(g) – 6 x  H f H 2 O (l) +  H f C 6 H 12 O 6 ∆H r = -(6 x -394) - (6 x -286) + (-1273) = +2807kJmol -1  H f C 6 H 12 O 6 -1273 6 x  H f H 2 O 6 x -286